Block at half its final speed

[rasen58]

A block of mass M starting from rest slides down a frictionless inclined plane of length L. When the block has attained 1/2 its final speed, the distance it has traveled along the plane is

 

choices: L/4, L/2, L/sqrt(2), 3L/4

 

I tried using the equation V²=v₀²+2ax and set v0 = 1/2v which becomes 1/4 v² to get 3/4 v² = 2ax and then 3a/8 v² = x

But I don't think I'm doing it right.

What should I be doing?

[imatfaal]

What's the relation between v and x in your equation? What shape would a graph be?

 

Hint: by saying look for the shape of a graph and the relation I mean that you concentrate on the fundamentals - the initial velocity is zero the acceleration is constant so you can basically ignore them

[studiot]

Three hints in the form of questions

 

At what distance travelled does the block reach its final speed?

 

In what direction do we take the velocity be for the purpose of the equations of motion?

Velocity is a vector do we need to resolve it in any particular directions?

Edited January 2, 2015 by studiot

[swansont]

A block of mass M starting from rest slides down a frictionless inclined plane of length L. When the block has attained 1/2 its final speed, the distance it has traveled along the plane is

 

choices: L/4, L/2, L/sqrt(2), 3L/4

 

I tried using the equation V²=v₀²+2ax and set v0 = 1/2v which becomes 1/4 v² to get 3/4 v² = 2ax and then 3a/8 v² = x

But I don't think I'm doing it right.

 

 

No, you aren't. v₀ isn't 1/2v, since it starts at rest. v₀ = 0

 

But you could use that equation to determine the speed at L, and then solve for half that speed. Using v₀ = 0, the equation becomes v_f² = 2ax

[rasen58]

Thanks

v² = 2ax

x = L

So then 1/2 of that final v = 1/2 sqrt(2aL)

And the equation for distance is now just x = 1/2 at² So we need to find a or t

So a = (v-v₀)/t and I then simplified that into t²=1/2 L/a

And plugged that back into the distance equation to get 1/4 L.

I think that's right?

Edited January 3, 2015 by rasen58

[swansont]

Thanks

v² = 2ax

x = L

So then 1/2 of that final v = 1/2 sqrt(2aL)

And the equation for distance is now just x = 1/2 at² So we need to find a or t

So a = (v-v₀)/t and I then simplified that into t²=1/2 L/a

And plugged that back into the distance equation to get 1/4 L.

I think that's right?

 

Or use the same equation. 1/2 v_f = sqrt(2aL)/2, as you say. So put that into the equation v² = 2ax (v₀ is still 0)and solve for x

 

2aL/4 = 2ax

 

x = L/4

 

Intuitively you might expect this, in terms of potential and kinetic energy. An object that doubles its speed has 4x the energy, so it has to travel 4x as far.

[andrea364499]

yes use that equation! 

1). set v to any letter, i chose v. set v₀ = 0 because it starts from rest. plug in L for Δx. solve for a. i got a = v² / 2L. 

2). using the V²=v₀²+2ax equation, set v = 1/2v, v₀ = 0, and a = v² / 2L. solve for Δx. i got Δx = L/4. 

 

hope this helps!

[Bufofrog]

3 minutes ago, andrea364499 said:

hope this helps!

He is going to get points off since his homework is 6 years late.

[swansont]

10 hours ago, Bufofrog said:

He is going to get points off since his homework is 6 years late.

They had already solved it, so it's all good

[Bufofrog]

12 minutes ago, swansont said:

They had already solved it, so it's all good

Well that's a relief.🙂